AP Physics 2 Practice Test 9 – 60 Questions with Answers

Test yourself with 60 original AP Physics 2 practice questions covering all 7 units of the course: thermodynamics, electric force, field, and potential, electric circuits, magnetism and electromagnetism, geometric optics, waves, sound, and physical optics, and modern physics. Every answer includes the full working. All questions are algebra-based (no calculus). Take g = 9.8 m/s2 unless stated otherwise.

Unit 9: Thermodynamics

Q1. A sample of gas stored in a small steel cylinder is initially at 100 kPa, occupies 12 L and is at 600 K. It is then compressed or expanded to 4 L and its temperature changes to 300 K. What is the final pressure of the gas?

  • (A) 300 kPa
  • (B) 150 kPa
  • (C) 75 kPa
  • (D) 225 kPa
Show Answer

Answer: B – P1V1 ÷ T1 = P2V2 ÷ T2, so P2 = P1 × (V1 ÷ V2) × (T2 ÷ T1) = 100 × (12 ÷ 4) × (300 ÷ 600) = 150 kPa.

Q2. 1200 J of heat is added to air in an engine cylinder while the gas does 700 J of work on its surroundings. What is the change in internal energy of the gas?

  • (A) 1900 J
  • (B) 500 J
  • (C) -500 J
  • (D) 1200 J
Show Answer

Answer: B – First law: ΔU = Q – Wby = 1200 – (700) = 500 J.

Q3. A gas in a expansion chamber expands at a constant pressure of 120 kPa from 4 L to 12 L. How much work is done by the gas?

  • (A) 960 J
  • (B) 1920 J
  • (C) 480 J
  • (D) None of the other options
Show Answer

Answer: A – W = PΔV = 120 kPa × (12 – 4) L = 120 × 8 = 960 J, since 1 kPa × 1 L = 1 J.

Q4. How much heat must be supplied to a 4 kg aluminium cooking pan of specific heat capacity 900 J/kg·K to raise its temperature by 15 K? (Assume no heat is lost to the surroundings.)

  • (A) 54,000 J
  • (B) 108,000 J
  • (C) 27,000 J
  • (D) 3,600 J
Show Answer

Answer: A – Q = mcΔT = 4 × 900 × 15 = 54,000 J.

Q5. A heat engine does 700 J of useful work in each cycle while absorbing 2000 J of heat from a hot reservoir. What is the efficiency of the engine?

  • (A) 65%
  • (B) 35%
  • (C) 70%
  • (D) 17.5%
Show Answer

Answer: B – Efficiency = W ÷ Qh = 700 ÷ 2000 = 0.35 = 35%.

Q6. An ideal Carnot engine operates between reservoirs at 600 K and 240 K. What is its maximum possible efficiency?

  • (A) 40%
  • (B) 30%
  • (C) 60%
  • (D) None of the other options
Show Answer

Answer: C – Carnot efficiency = 1 – Tc ÷ Th = 1 – 240 ÷ 600 = 1 – 0.4 = 0.6 = 60%.

Q7. The root-mean-square speed of the molecules in a gas sample is 600 m/s. If the absolute temperature of the gas is multiplied by 2.25 while the gas stays the same, what is the new root-mean-square speed?

  • (A) 1350 m/s
  • (B) 450 m/s
  • (C) 900 m/s
  • (D) 600 m/s
Show Answer

Answer: C – vrms is proportional to √T, so the speed is multiplied by √2.25 = 1.5. New vrms = 600 × 1.5 = 900 m/s.

Q8. Liquid water freezes into ice in a freezer. What happens to the entropy of the water itself?

  • (A) It increases
  • (B) It decreases
  • (C) It stays the same
  • (D) It becomes zero
Show Answer

Answer: B – Entropy measures disorder. Freezing orders the molecules into a crystal, so the entropy of the water falls.

Q9. What is the internal energy of 5 mol of a monatomic ideal gas at 300 K? (Use R = 8.31 J/mol·K.)

  • (A) 37395 J
  • (B) 18697.5 J
  • (C) 9348.75 J
  • (D) 12465 J
Show Answer

Answer: B – For a monatomic ideal gas, U = (3 ÷ 2)nRT = 1.5 × 5 × 8.31 × 300 = 18697.5 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two point charges of 6 μC and 2 μC are placed 0.6 m apart. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 0.6 N
  • (B) 0.15 N
  • (C) 1.2 N
  • (D) 0.3 N
Show Answer

Answer: D – F = kq1q2 ÷ r2 = 9.0 × 109 × (6 × 10-6) × (2 × 10-6) ÷ 0.62 = 0.3 N.

Q11. What is the magnitude of the electric field at a point 0.6 m from a point charge of 3 μC? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 150,000 N/C
  • (B) 37,500 N/C
  • (C) 75,000 N/C
  • (D) 300,000 N/C
Show Answer

Answer: C – E = kQ ÷ r2 = 9.0 × 109 × (3 × 10-6) ÷ 0.62 = 75,000 N/C.

Q12. What is the electric potential at a point 0.4 m from a point charge of 2 μC, taking potential at infinity to be zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 90,000 V
  • (B) 22,500 V
  • (C) 11,250 V
  • (D) 45,000 V
Show Answer

Answer: D – V = kQ ÷ r = 9.0 × 109 × (2 × 10-6) ÷ 0.4 = 45,000 V.

Q13. A charge of 6 μC is placed in a uniform electric field of 3000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.18 N
  • (B) 0.02 N
  • (C) 0 N
  • (D) 0.04 N
Show Answer

Answer: B – F = qE = (6 × 10-6) × 3000 = 0.02 N.

Q14. Two parallel metal plates are 0.1 m apart and have a potential difference of 150 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 3,000 V/m
  • (B) 750 V/m
  • (C) 1,500 V/m
  • (D) 15 V/m
Show Answer

Answer: C – E = V ÷ d = 150 ÷ 0.1 = 1,500 V/m.

Q15. How much work is done by the electric field when a charge of 2 μC moves through a potential drop of 600 V?

  • (A) 0.01 J
  • (B) None of the other options
  • (C) It cannot be determined from the data given
  • (D) 0 J
Show Answer

Answer: D – W = qΔV = (2 × 10-6) × 600 = 0 J.

Q16. Between two parallel plates, the upper plate is positive and the lower plate is negative. In which direction does the electric field between the plates point?

  • (A) From the lower plate toward the upper plate
  • (B) Parallel to the plates
  • (C) It is zero between the plates
  • (D) From the upper plate toward the lower plate
Show Answer

Answer: D – The electric field between parallel plates points from the positive plate toward the negative plate, and a negative charge accelerates opposite to the field direction.

Q17. The distance between two point charges is tripled, with both charges unchanged. How does the electrostatic force between them change?

  • (A) It becomes 1/3 of the original force
  • (B) It becomes 3 times the original force
  • (C) It becomes 1/9 of the original force
  • (D) It becomes 9 times the original force
Show Answer

Answer: C – F is proportional to 1 ÷ r2, so tripling r divides F by 9.

Q18. A alpha particle starting from rest is accelerated through a potential difference of 100 V. What kinetic energy does it gain, in electronvolts?

  • (A) 400 eV
  • (B) 100 eV
  • (C) 200 eV
  • (D) 800 eV
Show Answer

Answer: C – Kinetic energy gained = qV. An alpha particle carries two elementary charges, so KE = 2 × 100 = 200 eV.

Unit 11: Electric Circuits

Q19. A resistor carries a current of 4 A when a potential difference of 48 V is applied across it. What is its resistance?

  • (A) 12 Ω
  • (B) 192 Ω
  • (C) 0.083 Ω
  • (D) 52 Ω
Show Answer

Answer: A – R = V ÷ I = 48 ÷ 4 = 12 Ω.

Q20. Three resistors of 3 Ω, 9 Ω and 6 Ω are connected in series. What is their equivalent resistance?

  • (A) 9 Ω
  • (B) 28 Ω
  • (C) None of the other options
  • (D) 18 Ω
Show Answer

Answer: D – For series resistors, Req = R1 + R2 + R3 = 3 + 9 + 6 = 18 Ω.

Q21. Two resistors of 8 Ω and 8 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 16 Ω
  • (B) 4 Ω
  • (C) 8 Ω
  • (D) None of the other options
Show Answer

Answer: B – For two parallel resistors, Req = (R1R2) ÷ (R1 + R2) = (8 × 8) ÷ (8 + 8) = 64 ÷ 16 = 4 Ω.

Q22. A small motor operates at 9 V and draws 0.5 A. What power does it use?

  • (A) 18 W
  • (B) 9.5 W
  • (C) 4.5 W
  • (D) 0.056 W
Show Answer

Answer: C – P = VI = 9 × 0.5 = 4.5 W.

Q23. A battery of 20 V is connected across two resistors of 15 Ω and 5 Ω in series. What is the potential difference across the 15 Ω resistor?

  • (A) 5 V
  • (B) 30 V
  • (C) 7.5 V
  • (D) 15 V
Show Answer

Answer: D – The same current flows through both resistors, so V1 = V × R1 ÷ (R1 + R2) = 20 × 15 ÷ 20 = 15 V.

Q24. A 45 V battery is connected across two resistors in parallel. One branch has resistance 9 Ω. What current flows in that branch?

  • (A) 10 A
  • (B) 2.5 A
  • (C) 54 A
  • (D) 5 A
Show Answer

Answer: D – Each parallel branch has the full battery voltage, so I = V ÷ R = 45 ÷ 9 = 5 A.

Q25. A capacitor of capacitance 20 μF is charged to 30 V. How much energy is stored in it?

  • (A) 0.009 J
  • (B) 0.018 J
  • (C) 0.0045 J
  • (D) 0.9 J
Show Answer

Answer: A – E = 0.5CV2 = 0.5 × (20 × 10-6) × 302 = 0.5 × 2 × 10-5 × 900 = 0.009 J.

Q26. Two capacitors of 10 μF and 10 μF are connected in series. What is their equivalent capacitance?

  • (A) 20 μF
  • (B) 100 μF
  • (C) 10 μF
  • (D) 5 μF
Show Answer

Answer: D – Series: 1 ÷ Ceq = 1 ÷ 10 + 1 ÷ 10 = 2 ÷ 10, so Ceq = 5 μF.

Q27. A cell of emf 15 V and internal resistance 1.5 Ω supplies a current of 3 A. What is the terminal potential difference of the cell?

  • (A) 15 V
  • (B) 19.5 V
  • (C) 10.5 V
  • (D) 5.25 V
Show Answer

Answer: C – Terminal voltage V = emf – Ir = 15 – (3 × 1.5) = 15 – 4.5 = 10.5 V.

Unit 12: Magnetism and Electromagnetism

Q28. A particle with charge 10 μC moves at 3,000 m/s perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the particle?

  • (A) 0.01 N
  • (B) 0.12 N
  • (C) 0 N
  • (D) 0.02 N
Show Answer

Answer: A – F = qvB = (10 × 10-6) × 3,000 × 0.4 = 0.01 N.

Q29. A straight wire of length 1.2 m carries a current of 3 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the wire?

  • (A) 1.44 N
  • (B) 0.36 N
  • (C) 0.72 N
  • (D) 7.2 N
Show Answer

Answer: C – F = ILB = 3 × 1.2 × 0.2 = 0.72 N.

Q30. What is the magnitude of the magnetic field at a distance of 0.3 m from a long straight wire carrying a current of 9 A?

  • (A) 6 μT
  • (B) 12 μT
  • (C) 3 μT
  • (D) 60 μT
Show Answer

Answer: A – B = μ0I ÷ (2πr) = (2 × 10-7 × 9) ÷ 0.3 = 6 × 10-6 T = 6 μT.


Q31. A long solenoid has 400 turns per metre and carries a current of 2 A. What is the magnitude of the magnetic field inside the solenoid? (Use μ0 = 4π × 10-7 T·m/A and π = 3.14.)

  • (A) 1 mT
  • (B) 2.01 mT
  • (C) 0.5 mT
  • (D) 10.05 mT
Show Answer

Answer: A – B = μ0nI = 4π × 10-7 × 400 × 2 = 1 × 10-3 T = 1 mT.

Q32. A metal rod of length 0.75 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.6 T, with the rod perpendicular to both the field and its motion. What emf is induced between the ends of the rod?

  • (A) 1.8 V
  • (B) 0.45 V
  • (C) 0.9 V
  • (D) 9 V
Show Answer

Answer: C – Motional emf = BLv = 0.6 × 0.75 × 2 = 0.9 V.

Q33. The magnetic flux through a coil of 80 turns changes by 0.05 Wb in 0.2 s. What is the magnitude of the average induced emf in the coil?

  • (A) 40 V
  • (B) 10 V
  • (C) 2 V
  • (D) 20 V
Show Answer

Answer: D – Faraday’s law: emf = N × (Δflux ÷ time) = 80 × (0.05 ÷ 0.2) = 80 × 0.25 = 20 V.

Q34. A flat loop of area 0.25 m2 is placed in a uniform magnetic field of 0.8 T. The field makes an angle of 60° with the normal to the loop. What is the magnetic flux through the loop?

  • (A) 0.1 Wb
  • (B) 0.2 Wb
  • (C) 0.05 Wb
  • (D) None of the other options
Show Answer

Answer: A – Flux = BA cos θ = 0.8 × 0.25 × cos 60° = 0.2 × 0.5 = 0.1 Wb.

Q35. A bar magnet with its south pole facing a closed loop is pushed toward the loop. What magnetic pole does the side of the loop facing the magnet become?

  • (A) A south pole, so that it repels the approaching magnet
  • (B) A north pole, so that it attracts the approaching magnet
  • (C) It does not become a pole at all
  • (D) A south pole, so that it attracts the approaching magnet
Show Answer

Answer: A – Lenz’s law: an induced current always opposes the change that produced it. An approaching pole is repelled, a receding pole is attracted, and a falling magnet is slowed by the opposing induced currents.

Unit 13: Geometric Optics

Q36. An object is placed 36 cm in front of a concave dentist’s mirror of focal length 12 cm. How far from the mirror is the image formed?

  • (A) 36 cm
  • (B) 9 cm
  • (C) 24 cm
  • (D) 18 cm
Show Answer

Answer: D – Mirror equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 12 – 1 ÷ 36 = 0.06 cm-1, giving di = 18 cm.

Q37. An object is placed 40 cm in front of a converging lens of focal length 20 cm. How far from the lens is the image formed?

  • (A) 40 cm
  • (B) 80 cm
  • (C) 20 cm
  • (D) 60 cm
Show Answer

Answer: A – Lens equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 20 – 1 ÷ 40, giving di = 40 cm.

Q38. An object is 50 cm from a mirror and its image is formed 25 cm from the mirror. What is the lateral magnification of the image?

  • (A) 0.5
  • (B) -0.5
  • (C) -1
  • (D) -0.25
Show Answer

Answer: B – Magnification m = -di ÷ do = -25 ÷ 50 = -0.5. The negative sign shows the image is inverted.

Q39. A ray of light travels from a medium of refractive index 1.33 into a medium of refractive index 1 at an angle of incidence of 30°. What is the angle of refraction, to the nearest 0.1°?

  • (A) 51.7°
  • (B) 41.7°
  • (C) 31.7°
  • (D) 48.3°
Show Answer

Answer: B – Snell law: n1 sin θ1 = n2 sin θ2, so sin θ2 = (1.33 × sin 30°) ÷ 1 = 0.66. Hence θ2 = 41.7°.

Q40. What is the critical angle for light travelling from a plastic of refractive index 1.25 into air, to the nearest 0.1°?

  • (A) 63.1°
  • (B) 43.1°
  • (C) 36.9°
  • (D) 53.1°
Show Answer

Answer: D – At the critical angle, sin θc = n2 ÷ n1 = 1 ÷ 1.25 = 0.8, so θc = 53.1°.

Q41. Light travels at 2.4 × 108 m/s in a transparent material. What is the refractive index of the material? (Take c = 3.00 × 108 m/s.)

  • (A) 2.5
  • (B) 1.25
  • (C) 0.62
  • (D) 0.6
Show Answer

Answer: B – n = c ÷ v = (3.00 × 108) ÷ (2.4 × 108) = 1.25.

Q42. Light of wavelength 625 nm in air enters a medium of refractive index 1.25. What is its wavelength in the medium?

  • (A) 500 nm
  • (B) 625 nm
  • (C) 1000 nm
  • (D) 250 nm
Show Answer

Answer: A – Wavelength in a medium: λ = λ0 ÷ n = 625 ÷ 1.25 = 500 nm. The frequency stays the same, so the wavelength shortens.

Q43. An object is placed beyond twice the focal length of a converging lens. Which description best fits the image formed?

  • (A) Virtual, upright and magnified
  • (B) Real, upright and diminished
  • (C) Virtual, inverted and diminished
  • (D) Real, inverted and diminished
Show Answer

Answer: D – Ray diagrams show: a concave mirror gives a real, inverted, diminished image beyond the centre of curvature and a virtual, upright, magnified image inside the focal point; convex mirrors and diverging lenses always give virtual, upright, diminished images; a converging lens with the object beyond 2F gives a real, inverted, diminished image.

Unit 14: Waves, Sound, and Physical Optics

Q44. A wave has frequency 15 Hz and wavelength 10 m. What is its speed?

  • (A) 300 m/s
  • (B) 75 m/s
  • (C) 25 m/s
  • (D) 150 m/s
Show Answer

Answer: D – v = fλ = 15 × 10 = 150 m/s.

Q45. A string of length 2.5 m is fixed at both ends. Waves travel along it at 200 m/s. What is the frequency of its fundamental?

  • (A) 80 Hz
  • (B) 20 Hz
  • (C) 40 Hz
  • (D) None of the other options
Show Answer

Answer: C – For a string fixed at both ends, fn = nv ÷ (2L) = 1 × 200 ÷ (2 × 2.5) = 40 Hz.

Q46. An air column of length 1.7 m is closed at one end. Taking the speed of sound as 340 m/s, what is its fundamental frequency?

  • (A) 100 Hz
  • (B) 50 Hz
  • (C) 25 Hz
  • (D) 200 Hz
Show Answer

Answer: B – For a pipe closed at one end, f = v ÷ (4L) = 340 ÷ (4 × 1.7) = 50 Hz.

Q47. Two tuning forks of frequencies 300 Hz and 293 Hz are sounded together. What is the beat frequency heard?

  • (A) 7 Hz
  • (B) 14 Hz
  • (C) 9 Hz
  • (D) 593 Hz
Show Answer

Answer: A – Beat frequency = |f1 – f2| = |300 – 293| = 7 Hz.

Q48. In a double-slit experiment, light of wavelength 550 nm passes through slits 0.25 mm apart and forms fringes on a screen 2 m away. How far from the central bright fringe is the first bright fringe, in millimetres?

  • (A) 8.8 mm
  • (B) 2.2 mm
  • (C) 4.4 mm
  • (D) 44 mm
Show Answer

Answer: C – Fringe position y = mλL ÷ d = 1 × (550 × 10-9) × 2 ÷ (0.25 × 10-3) = 4.4 × 10-3 m = 4.4 mm.

Q49. A sound source emitting 400 Hz moves at 68 m/s away from a stationary observer. Taking the speed of sound as 340 m/s, what frequency does the observer hear, to the nearest hertz?

  • (A) 333 Hz
  • (B) 373 Hz
  • (C) 293 Hz
  • (D) 400 Hz
Show Answer

Answer: A – Doppler shift for a moving source: f’ = f × v ÷ (v + vs) = 400 × 340 ÷ 408 = 333 Hz to the nearest hertz.

Q50. A listener moves to four times the original distance from a point source of sound. How does the sound intensity change?

  • (A) It becomes 1/4 of the original intensity
  • (B) It becomes 1/8 of the original intensity
  • (C) It becomes 1/16 of the original intensity
  • (D) It stays the same
Show Answer

Answer: C – I proportional to 1 ÷ r2, so multiplying r by 4 divides I by 16.

Q51. A diffraction grating has 300 lines per millimetre. Light of wavelength 600 nm is incident normally on it. What is the angle of the first-order maximum, to the nearest 0.1°?

  • (A) 20.4°
  • (B) 0.4°
  • (C) 10.4°
  • (D) 79.6°
Show Answer

Answer: C – Grating spacing d = 1 ÷ 300 mm = 3333.33 nm. For the first order, d sin θ = λ, so sin θ = 600 ÷ 3333.33 = 0.18, giving θ = 10.4°.

Unit 15: Modern Physics

Q52. What is the energy, in electronvolts, of a photon of wavelength 310 nm? (Use hc = 1240 eV·nm.)

  • (A) 8 eV
  • (B) 2 eV
  • (C) 6 eV
  • (D) 4 eV
Show Answer

Answer: D – E = hc ÷ λ = 1240 ÷ 310 = 4 eV.

Q53. Light of photon energy 4.5 eV falls on a metal surface whose work function is 3.1 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 1.4 eV
  • (B) 7.6 eV
  • (C) 4.5 eV
  • (D) 2.8 eV
Show Answer

Answer: A – Einstein photoelectric equation: Kmax = hf – work function = 4.5 – 3.1 = 1.4 eV.

Q54. A metal has a work function of 1.5 eV. What is its threshold frequency, to two decimal places in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s and 1 eV = 1.60 × 10-19 J.)

  • (A) 7.24 × 1014 Hz
  • (B) 1.81 × 1014 Hz
  • (C) 5.43 × 1014 Hz
  • (D) 3.62 × 1014 Hz
Show Answer

Answer: D – Threshold: hf0 = work function = 1.5 × 1.60 × 10-19 J = 2.4 × 10-19 J. So f0 = (2.4 × 10-19) ÷ (6.63 × 10-34) = 3.62 × 1014 Hz.

Q55. An electron moves at 1.5 × 106 m/s. What is its de Broglie wavelength, in nanometres? (Use h = 6.63 × 10-34 J·s and electron mass 9.11 × 10-31 kg.)

  • (A) 0.97 nm
  • (B) 0.49 nm
  • (C) 0.24 nm
  • (D) 4.85 nm
Show Answer

Answer: B – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 1.5 × 106) = 0.49 × 10-9 m = 0.49 nm.

Q56. A hydrogen atom makes a transition from energy level n = 3 to n = 1. How much energy is released as a photon, in electronvolts? (Use En = -13.6 ÷ n2 eV.)

  • (A) 24.18 eV
  • (B) 12.09 eV
  • (C) 6.04 eV
  • (D) 1.51 eV
Show Answer

Answer: B – Energy released = E3 – E1 in magnitude = 13.6 × (1 ÷ 12 – 1 ÷ 32) = 13.6 × 0.89 = 12.09 eV.

Q57. A radioactive sample initially contains 200 g of an isotope. After 3 half-lives, how much of the isotope remains?

  • (A) 50 g
  • (B) 25 g
  • (C) 12.5 g
  • (D) 175 g
Show Answer

Answer: B – After each half-life half remains, so after 3 half-lives the fraction left is (1 ÷ 2)3 = 1 ÷ 8. Remaining mass = 200 ÷ 8 = 25 g.

Q58. In a nuclear reaction the total mass decreases by 0.03 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 55.89 MeV
  • (B) 27.95 MeV
  • (C) 13.97 MeV
  • (D) 279.45 MeV
Show Answer

Answer: B – E = Δmc2 = 0.03 × 931.5 = 27.95 MeV.

Q59. A clock on a spacecraft moving at 0.6c ticks 20 s as measured in the spacecraft frame. How much time passes on Earth while the spacecraft clock ticks 20 s?

  • (A) 20 s
  • (B) 50 s
  • (C) 25 s
  • (D) 40 s
Show Answer

Answer: C – For v = 0.6c, the Lorentz factor is 1.25. Dilated time = 1.25 × 20 = 25 s.

Q60. A spacecraft of proper length 90 m travels past Earth at 0.8c. What length does an observer on Earth measure for the spacecraft?

  • (A) 54 m
  • (B) 90 m
  • (C) 108 m
  • (D) 45 m
Show Answer

Answer: A – Length contraction: L = L0 ÷ Lorentz factor = 90 ÷ 1.67 = 54 m.

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